this is a linear second order homogeneous differential equation that involves the first second and 0th derivatives of a function. Therefore the solution to this differential equation will have a solution of the form e^(rx) where r is uniquely determined by the coefficients of the 1st 2nd and 0th derivatives of y.

yes, sort of. you look at the equation of this form. ay"+by′+cy=0, and you take the coefficients a, b, and c and then plug them into a corresponding 2nd degree polynomial in terms of r. so for ay"+by′+cy=0, you have \[ar^2+br+c =0\] solving this polynomial will give you the values of r for which you will have solutions of the form \[e ^{rx}\]

so yea, set up your 2nd degree polynomial using the coefficients from the initial equation:
y"+6y'+8y=0
\[r^2 +6r+8=0\]
like dumbcow said, the left side of this equation can be easily factored:
\[(r+4)(r+2)=0\]
the solutions to this equation are therefore r=-4 and r=-2
so therefore we have two linearly independent solutions
\[y = e ^{-4x}\] and \[y = e ^{-2x}\]
both of these equations are solutions to the differential equation y"+6y'+8y=0. Every possible linear combination of these equations is as well, ad since they are linearly independent, the sum of these equations is also a solution to y"+6y'+8y=0. Try all of these out use \[y = e ^{-4x}\], \[y = e ^{-2x}\], \[y = 4e ^{-4x}\] and \[y = e ^{-4x}+e ^{-2x}\]

you'll find that they will all satisfy y"+6y'+8y=0. So in order to provide a general equation that expresses all possible solutions we write \[y=Ke ^{-4x}+Ce ^{-2x}\] where K and C are arbitrary constants